This is a basic question, but I don't know the answer. Suppose $M$ is an $R$-module, considered as a complex concentrated in degree $0$. Let $F^*$ be an complex consisting of free modules. Recall that $F^*$ is called a free resolution of $M$ if there is a quasi-isomorphism $F^*\to M$.

Now suppose that $M\cong F^*$ in the derived category of $R$-modules. Does this mean that $F^*$ is a free resolution of $M$? I guess not, since maybe the map goes "in the wrong direction", but I cannot think of an example.

  • $\begingroup$ I think you need to ask that $F_\bullet$ is a dg-projective complex (i.e. a complex of projective modules such that every map from $F_\bullet$ to an exact complex is nullhomotopic). For example, all bounded below complexes of projective modules are dg-projective (Warning: I'm using homological grading) $\endgroup$ – Denis Nardin May 7 '18 at 13:49

Maybe you need to take a truncation.

For a counterexample,consider $Z_4-Mod$,$0\rightarrow Z_4/2Z_4\rightarrow Z_4\rightarrow Z_4\rightarrow …$ is injective resolution of $Z_4/2Z_4$,how do you think this as projective resolution?

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